How many different ways to play doubles tennis ? : GMAT Problem Solving (PS)
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# How many different ways to play doubles tennis ?

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How many different ways to play doubles tennis ? [#permalink]

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12 Jul 2011, 01:20
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A group of 8 friends want to play doubles tennis. How many different ways can the group be divided into 4 teams of 2 people?

A. 420
B. 2520
C. 168
D. 90
E. 105
[Reveal] Spoiler: OA
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Re: How many different ways to play doubles tennis ? [#permalink]

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12 Jul 2011, 01:26
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Alchemist1320 wrote:
A group of 8 friends want to play doubles tennis. How many different ways can the group be divided into 4 teams of 2 people?

A. 420
B. 2520
C. 168
D. 90
E. 105

Formula = The number of ways in which MN different items can be divided equally into M groups, each containing N objects and the order of the groups is not important important is (mn)!/n^m*m!

so its 8!/(2^4*4!) = 105 = E
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Re: How many different ways to play doubles tennis ? [#permalink]

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12 Jul 2011, 02:42
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Expert's post
Alchemist1320 wrote:
A group of 8 friends want to play doubles tennis. How many different ways can the group be divided into 4 teams of 2 people?

A. 420
B. 2520
C. 168
D. 90
E. 105

From the question, we can say that the teams are not distinct i.e. we don't have team A, team B etc. But let's solve this question by making first team, second team, third team and fourth team. Later we will adjust the answer.
Out of 8 people, how can you make the first team? In 8C2 ways.
Out of 6 people, how can you make the second team? In 6C2 ways.
Out of 4 people, how can you make the third team? In 4C2 ways.
Out of 2 people, how can you make the fourth team? In 2C2 ways.
How can you make the 4 teams?
8C2 *6C2*4C2*2C2

But here, we have considered the 4 teams to be distinct. We called them 'first team', 'second team' etc.
So we need to divide the result by 4! to 'un-arrange' them.

You get: 8C2 *6C2*4C2*2C2/4! = 8*7*6*5*4*3*2*1/16*4! = 105
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Manager Joined: 14 Feb 2011 Posts: 69 Followers: 1 Kudos [?]: 1 [0], given: 2 Re: How many different ways to play doubles tennis ? [#permalink] ### Show Tags 08 Aug 2011, 06:16 sudhir18n wrote: Alchemist1320 wrote: A group of 8 friends want to play doubles tennis. How many different ways can the group be divided into 4 teams of 2 people? A. 420 B. 2520 C. 168 D. 90 E. 105 Formula = The number of ways in which MN different items can be divided equally into M groups, each containing N objects and the order of the groups is not important important is (mn)!/n^m*m! so its 8!/(2^4*4!) = 105 = E super! thanks for formula! Manager Status: Do till 740 :) Joined: 13 Jun 2011 Posts: 113 Concentration: Strategy, General Management GMAT 1: 460 Q35 V20 GPA: 3.6 WE: Consulting (Computer Software) Followers: 1 Kudos [?]: 11 [0], given: 19 Re: How many different ways to play doubles tennis ? [#permalink] ### Show Tags 14 Sep 2011, 13:20 karishma, Quote: But here, we have considered the 4 teams to be distinct. We called them 'first team', 'second team' etc. So we need to divide the result by 4! to 'un-arrange' them. Can you please explain this? I understand that we divide the slots! to remove identical stuff but here how does it make sense? Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7187 Location: Pune, India Followers: 2168 Kudos [?]: 14025 [2] , given: 222 Re: How many different ways to play doubles tennis ? [#permalink] ### Show Tags 14 Sep 2011, 21:07 2 This post received KUDOS Expert's post shankar245 wrote: Can you please explain this? I understand that we divide the slots! to remove identical stuff but here how does it make sense? This is the logic behind this step: Say there are 4 boys: A, B, C, D There are two ways of splitting them in two groups. Method I The two groups can be made in the following ways 1. AB and CD 2. AC and BD 3. AD and BC The groups are not named/distinct. You have 4 boys in front of you and you split them in 2 groups and do not name the groups. There are 3 total ways of doing this. Method II On the other hand, I could put them in two distinct groups in the following ways 1. Group1: AB, Group2: CD 2. Group1: CD, Group2: AB (If you notice, this is the same as above, just that now AB is group 2) 3. Group1: AC, Group2: BD 4. Group1: BD, Group2: AC 5. Group1: AD, Group2: BC 6. Group1: BC, Group2: AD Here I have to put them in two different groups, group 1 and group 2. AB and CD is not just one way of splitting them. AB could be assigned to group 1 or group 2 so there are 2 cases. In this case, every 'way' we get above will have two possibilities so total number of ways will be twice. So there will be 6 total ways. Here since the groups are not distinct but 8C2 * 6C2 * 4C2 * 2C2 makes them distinct (we say, select the FIRST group in 8C2 ways, SECOND group in 6C2 ways etc), we need to divide by 4!. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: How many different ways to play doubles tennis ? [#permalink]

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14 Sep 2011, 21:41
thanks karishma for that detailed explanation
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Re: How many different ways to play doubles tennis ? [#permalink]

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15 Sep 2011, 07:57
Thanks Sudhir , I did not (know) the formula .. and was coming up with 8C2 + 6C2+ 4C2 + 2C2 which is obviously wrong. After I read the formula, it came back to me from school days!!

Thanks Karishma, for the detailed explanation it was really helpful.
Re: How many different ways to play doubles tennis ?   [#permalink] 15 Sep 2011, 07:57
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# How many different ways to play doubles tennis ?

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